Answer to Question #131390 in Optics for Reeshabh Kumar

Question

When a liquid is introduced between the length and the plate in the NEWTON'S ring Apparatus the diameter of the 15th ring changes from 2cm to 1.8 CM. The refractive index of a liquid is

Answer 1

Solution

Given data in the question

Order of newton ring (n) =15

Before introducing any liquid between

the length and the plate in the Newton's ring Apparatus there was air so it's Diameter $(D_n) _{air}=2 cm$

$=2\times10^{-2}m$

After introduced liquid diameter

$(D_n) _{liquid}=1.8cm$

$=1.8\times10^{-2}m$

Refractive index of this liquid can be written as

$\mu=\frac{(D_n) ^2_{air}}{(D_n)^2 _{liquid}}$

$\mu=\frac{(2\times10^{-2}) ^2}{(1.8\times10^{-2}) ^2}$

$\mu=\frac{4}{3.24}$

$\fcolorbox{red}{yellow}{$\mu=1.23$}$

Therefore refractive index of liquid is 1.23 .